Find points at a given distance on a line of given slope

Given the co-ordinates of a 2-dimensional point p(x₀, y₀). Find the points at a distance L away from it, such that the line formed by joining these points has a slope of M.

Examples:

Input : p = (2, 1)
        L = sqrt(2)
        M = 1
Output :3, 2
        1, 0
Explanation:
The two points are sqrt(2) distance away 
from the source and have the required slope
m = 1.

Input : p = (1, 0)
        L = 5
        M = 0
Output : 6, 0
        -4, 0

We need to find two points that are L distance from given point, on a line with slope M.
The idea has been introduced in below post.
Find Corners of Rectangle using mid points

Based on the input slope, the problem can be classified into 3 categories.

If slope is zero, we just need to adjust the x coordinate of the source point
If slope is infinite, the we need to adjust the y coordinate
For other values of slope, we can use the following equations to find the points

$Given \ that \ the \ point (x, y) \ is \ at \ distance \ I \ away \ from \ (x_0, y_0) \newline \newline (y-y_0)^{2} + (x-x_0)^{2}= l^{2} \newline \newline Also \ as \ the \ line \ that \ passes \ through \ (x, y) \ and \ (x0, y0) \ satisfies \newline \newline \frac{y-y_0}{x-x_0}= m \newline \newline Rearranging \ we \ get \newline y=y_0+m*(x-x_0) \newline \newline Putting \ the \ values \ in \ first \ equation \newline \newline m^2.(x-x_0)^2+(x-x_0)^2=l^2 \newline \newline Hence, \ we \ have \newline \newline x=x_0\pm l.\sqrt{\frac{1}{1+m^2}} \newline \newline y=y_0 \pm m.l.\sqrt{\frac{1}{1+m^2}}$

Now using the above formula we can find the required points.

C++

// C++ program to find the points on a line of
// slope M at distance L
#include <bits/stdc++.h>
using namespace std;
 
// structure to represent a co-ordinate
// point
struct Point {
 
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a, y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
void printPoints(Point source, float l, 
                                 int m)
{
    // m is the slope of line, and the 
    // required Point lies distance l 
    // away from the source Point
    Point a, b;
 
    // slope is 0
    if (m == 0) {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // if slope is infinte
    else if (m == std::numeric_limits<float>
                                 ::max()) {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    } 
    else {
        float dx = (l / sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // print the first Point
    cout << a.x << ", " << a.y << endl;
 
    // print the second Point
    cout << b.x << ", " << b.y << endl;
}
 
// driver function
int main()
{
    Point p(2, 1), q(1, 0);
    printPoints(p, sqrt(2), 1);
    cout << endl;
    printPoints(q, 5, 0);
    return 0;
}

Java

// Java program to find the points on  
// a line of slope M at distance L 
class GFG{
 
// Class to represent a co-ordinate
// point
static class Point 
{
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source, 
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinte
    else if (Double.isInfinite(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    } 
    else
    {
        float dx = (float)(l / Math.sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    System.out.println(a.x + ", " + a.y);
 
    // Print the second Point
    System.out.println(b.x + ", " + b.y);
}
 
// Driver code
public static void main(String[] args)
{
    Point p = new Point(2, 1), 
          q = new Point(1, 0);
    printPoints(p, (float)Math.sqrt(2), 1);
     
    System.out.println();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Rajnis09

C#

// C# program to find the points on  
// a line of slope M at distance L 
using System;
 
class GFG{
 
// Class to represent a co-ordinate
// point
public class Point 
{
    public float x, y;
     
    public Point()
    {
        x = y = 0;
    }
     
    public Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source, 
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinte
    else if (Double.IsInfinity(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    } 
    else
    {
        float dx = (float)(l / Math.Sqrt(
                           1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    Console.WriteLine(a.x + ", " + a.y);
 
    // Print the second Point
    Console.WriteLine(b.x + ", " + b.y);
}
 
// Driver code
public static void Main(String[] args)
{
    Point p = new Point(2, 1), 
          q = new Point(1, 0);
           
    printPoints(p, (float)Math.Sqrt(2), 1);
     
    Console.WriteLine();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Amit Katiyar

Output:

3, 2
1, 0

6, 0
-4, 0

This article is contributed by Ashutosh Kumar 😀 If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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My Personal Notes

C++

Java

C#

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